What is the probability of the sum of four dice being 22?

There aren't too many to just count.

Permutations of $6+6+6+4$: $\binom41=4$

Permutations of $6+6+5+5$: $\binom42=6$

These are the only options, so your numerator must be $4+6=10$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The number of configurations that satisfies "the sum is $\ds{22}$" is given by:

\begin{align} X & = \sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6}\sum_{d_{4} = 1}^{6} \bracks{z^{22}}z^{d_{1} + d_{2} + d_{3} + d_{4}} = \bracks{z^{22}}\pars{\sum_{d = 1}^{6}z^{d}}^{4} = \bracks{z^{22}}\pars{z\,{z^{6} - 1 \over z - 1}}^{4} \\[5mm] & = \bracks{z^{18}}{1 - 4z^{6} + 6z^{12} - 4z^{18} + z^{24}\over \pars{1 - z}^{4}} \\[5mm] & = \bracks{z^{18}}\pars{1 - z}^{-4} - 4\bracks{z^{12}}\pars{1 - z}^{-4} + 6\bracks{z^{6}}\pars{1 - z}^{-4} - 4\bracks{z^{0}}\pars{1 - z}^{-4} \\[5mm] & = {-4 \choose 18}\pars{-1}^{18} - 4{-4 \choose 12}\pars{-1}^{12} + 6{-4 \choose 6}\pars{-1}^{6} - 4 = {21 \choose 18} - 4{15 \choose 12} + 6{9 \choose 6} - 4 \\[5mm] & = 1330 -4 \times 455 + 6 \times 84 - 4 = \bbx{10} \end{align}

The bad combinations criteria is atleast one $x_i \geq 7$.

The number of bad combinations when:

1. One of $x_i$ is forced to be greater than or equal to $7$ is $\binom{4}{1}\binom{12 + 4 - 1}{12} = 1820$.

2. Two of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{2}\binom{6+4-1}{6} = 504$.

3. Three of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{3}\binom{0+4-1}{0} = 4$.

4. Four of $x_i$'s are forced to be greater than or equal to $7$ is $0$.

So, total bad combinations $= 1820 - 504 + 4 - 0 = 1320$

I used $n(\cup_{i=1}^{4}A_i) = \sum_{i=1}^{4}n(A_i) - \sum_{i,j, i\neq j}n(A_i\cap A_j) + \ldots$

So, possible combinations $= 1330 - 1320 = 10$.

In order for the sum to equal 22, either three dice equal $6$ and one equals $4$, or two dice equal $6$ and two dice equal $5$. The number of valid outcomes thus equals:

$${4 \choose 1} + {4 \choose 2} = 4 + 6 = 10$$

As such, the probability of the four dice having a sum of $22$ equals:

$$\frac{{4 \choose 1} + {4 \choose 2}}{6^4} = \frac{10}{1296} = \frac{5}{648} \approx 0.00772$$

@expiTTp1z0 has addressed where you made your error.

I am going to show you how you can reduce the given problem a simpler one by using symmetry.

You wish to find the number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 = 22 \tag{1}$$ in positive integers not exceeding $6$. Since $x_k$, $1 \leq k \leq 4$, is a positive integer satisfying $x_k \leq 6$, then $y_k = 7 - x_k$ is also a positive integer not exceeding $6$. Substituting $7 - y_k$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields \begin{align*} 7 - y_1 + 7 - y_2 + 7 - y_3 + 7 - y_4 & = 22\\ -y_1 - y_2 - y_3 - y_4 & = -6\\ y_1 + y_2 + y_3 + y_4 & = 6 \tag{2} \end{align*} Equation 2 is an equation in the positive integers when the given restrictions are imposed. A particular solution of equation corresponds to placing an addition sign in three of the five spaces between successive ones in a row of six ones. For instance, $$1 + 1 1 + 1 + 1 1$$ corresponds to the solution $y_1 = 1$, $y_2 = 2$, $y_3 = 1$, and $y_4 = 2$ (or $x_1 = 6$, $x_2 = 5$, $x_3 = 6$, $x_4 = 5$) while $$1 1 1 + 1 + 1 + 1$$ corresponds to the solution $y_1 = 3$, $y_2 = y_3 = y_4 = 1$ (or $x_1 = 4$, $x_2 = x_3 = x_4 = 6$). Thus, the number of solutions of equation 2 in the positive integers is the number of ways of selecting which three of the five spaces between successive ones in a row of six ones will be filled by addition signs, which is $$\binom{5}{3} = 10$$

Note: If you prefer to work in the nonnegative integers, then we wish to find the number of solutions of the equation $$x_1' + x_2' + x_3' + x_4' = 18 \tag{$1'$}$$ subject to the restrictions that $x_k' = x_k - 1 \leq 6 - 1 = 5$ for $1 \leq k \leq 4$. Since $x_k'$, $1 \leq k \leq 5$, is a nonnegative integer not exceeding $5$, so is $y_k' = 5 - x_k'$. Substituting $5 - y_k'$ for $x_k'$ in equation 1' and simplifying yields $$y_1' + y_2' + y_3' + y_4' = 2 \tag{$2'$}$$ which is an equation in the nonnegative integers. A particular solution corresponds to the insertion of three addition signs in a row of two ones. For instance, $$+ + + 1 1$$ corresponds to the solution $y_1 = y_2 = y_3 = 0$, $y_4 = 2$ (or $x_1 = x_2 = x_3 = 6$, $x_4 = 4$), while $$1 + 1 + +$$ corresponds to the solution $y_1 = y_2 = 1$, $y_3 = y_4 = 0$ (or $x_1 = x_2 = 5$, $x_3 = x_4 = 6$). Thus, the number of solutions of equation 2 in the nonnegative integers is $$\binom{2 + 3}{3} = \binom{5}{3}$$ since we must choose which three of the five positions (two ones and three addition signs) will be filled with addition signs.